Write a removeDuplicates() function which takes a list and deletes any duplicate nodes from the list. The list is not sorted.
For example if the linked list is 12->11->12->21->41->43->21 then removeDuplicates() should convert the list to 12->11->21->41->43.
METHOD 1 (Using two loops)
This is the simple way where two loops are used. Outer loop is used to pick the elements one by one and inner loop compares the picked element with rest of the elements.
Java Programming:
// Java program to remove duplicates from unsorted
// linked list
class LinkedList {
static Node head;
static class Node {
int data;
Node next;
Node(int d) {
data = d;
next = null;
}
}
/* Function to remove duplicates from an
unsorted linked list */
void remove_duplicates() {
Node ptr1 = null, ptr2 = null, dup = null;
ptr1 = head;
/* Pick elements one by one */
while (ptr1 != null && ptr1.next != null) {
ptr2 = ptr1;
/* Compare the picked element with rest
of the elements */
while (ptr2.next != null) {
/* If duplicate then delete it */
if (ptr1.data == ptr2.next.data) {
/* sequence of steps is important here */
dup = ptr2.next;
ptr2.next = ptr2.next.next;
System.gc();
} else /* This is tricky */ {
ptr2 = ptr2.next;
}
}
ptr1 = ptr1.next;
}
}
void printList(Node node) {
while (node != null) {
System.out.print(node.data + " ");
node = node.next;
}
}
public static void main(String[] args) {
LinkedList list = new LinkedList();
list.head = new Node(10);
list.head.next = new Node(12);
list.head.next.next = new Node(11);
list.head.next.next.next = new Node(11);
list.head.next.next.next.next = new Node(12);
list.head.next.next.next.next.next = new Node(11);
list.head.next.next.next.next.next.next = new Node(10);
System.out.println("Linked List before removing duplicates : \n ");
list.printList(head);
list.remove_duplicates();
System.out.println("");
System.out.println("Linked List after removing duplicates : \n ");
list.printList(head);
}
}
// This code has been contributed by Mayank Jaiswal
Output :
Linked list before removing duplicates:
10 12 11 11 12 11 10
Linked list after removing duplicates:
10 12 11
Time Complexity: O(n^2)
METHOD 2 (Use Sorting)
In general, Merge Sort is the best suited sorting algorithm for sorting linked lists efficiently.
1) Sort the elements using Merge Sort. We will soon be writing a post about sorting a linked list. O(nLogn)
2) Remove duplicates in linear time using the algorithm for removing duplicates in sorted Linked List. O(n)
Please note that this method doesn’t preserve the original order of elements.
Time Complexity: O(nLogn)
METHOD 3 (Use Hashing)
We traverse the link list from head to end. For every newly encountered element, we check whether it is in the hash table: if yes, we remove it; otherwise we put it in the hash table.
C++ Programming:
/* Program to remove duplicates in an unsorted
linked list */
#include<bits/stdc++.h>
using namespace std;
/* A linked list node */
struct Node
{
int data;
struct Node *next;
};
// Utility function to create a new Node
struct Node *newNode(int data)
{
Node *temp = new Node;
temp->data = data;
temp->next = NULL;
return temp;
}
/* Function to remove duplicates from a
unsorted linked list */
void removeDuplicates(struct Node *start)
{
// Hash to store seen values
unordered_set<int> seen;
/* Pick elements one by one */
struct Node *curr = start;
struct Node *prev = NULL;
while (curr != NULL)
{
// If current value is seen before
if (seen.find(curr->data) != seen.end())
{
prev->next = curr->next;
delete (curr);
}
else
{
seen.insert(curr->data);
prev = curr;
}
curr = prev->next;
}
}
/* Function to print nodes in a given linked list */
void printList(struct Node *node)
{
while (node != NULL)
{
printf("%d ", node->data);
node = node->next;
}
}
/* Driver program to test above function */
int main()
{
/* The constructed linked list is:
10->12->11->11->12->11->10*/
struct Node *start = newNode(10);
start->next = newNode(12);
start->next->next = newNode(11);
start->next->next->next = newNode(11);
start->next->next->next->next = newNode(12);
start->next->next->next->next->next =
newNode(11);
start->next->next->next->next->next->next =
newNode(10);
printf("Linked list before removing duplicates : \n");
printList(start);
removeDuplicates(start);
printf("\nLinked list after removing duplicates : \n");
printList(start);
return 0;
}
Output :
Linked list before removing duplicates:
10 12 11 11 12 11 10
Linked list after removing duplicates:
10 12 11
Time Complexity: O(n) on average (assuming that hash table access time is O(1) on average).